The value of `Delta = |(1^(2),2^(2),3^(2)),(2^(2),3^(2),4^(2)),(3^(2),4^(2),5^(2))|`, is

Options

A. 8
B. `-8'
C. 400
D. 1

Answer

To find the value of the determinant \( \Delta = \begin{vmatrix} 1^2 & 2^2 & 3^2 \\ 2^2 & 3^2 & 4^2 \\ 3^2 & 4^2 & 5^2 \end{vmatrix} \), we will follow these steps:

### Step 1: Write the Determinant
First, we will write the determinant with the actual values:
\[ \Delta = \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{vmatrix} \]

### Step 2: Apply Row Operations
We can simplify the determinant by performing row operations. We will subtract the second row from the third row:
\[ R_3 \leftarrow R_3 - R_2 \]
This gives us:
\[ \Delta = \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 5 & 7 & 9 \end{vmatrix} \]

### Step 3: Apply Another Row Operation
Next, we will subtract the first row from the third row:
\[ R_3 \leftarrow R_3 - R_1 \]
This results in:
\[ \Delta = \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 4 & 3 & 0 \end{vmatrix} \]

### Step 4: Expand the Determinant
Now we will expand the determinant along the first row:
\[ \Delta = 1 \cdot \begin{vmatrix} 9 & 16 \\ 3 & 0 \end{vmatrix} - 4 \cdot \begin{vmatrix} 4 & 16 \\ 4 & 0 \end{vmatrix} + 9 \cdot \begin{vmatrix} 4 & 9 \\ 4 & 3 \end{vmatrix} \]

### Step 5: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:
1. \( \begin{vmatrix} 9 & 16 \\ 3 & 0 \end{vmatrix} = (9 \cdot 0) - (16 \cdot 3) = -48 \)
2. \( \begin{vmatrix} 4 & 16 \\ 4 & 0 \end{vmatrix} = (4 \cdot 0) - (16 \cdot 4) = -64 \)
3. \( \begin{vmatrix} 4 & 9 \\ 4 & 3 \end{vmatrix} = (4 \cdot 3) - (9 \cdot 4) = 12 - 36 = -24 \)

### Step 6: Substitute Back into the Expansion
Substituting these values back into the expansion gives:
\[ \Delta = 1 \cdot (-48) - 4 \cdot (-64) + 9 \cdot (-24) \]
\[ = -48 + 256 - 216 \]

### Step 7: Simplify the Expression
Now we simplify:
\[ \Delta = -48 + 256 - 216 = -48 + 40 = -8 \]

### Final Answer
Thus, the value of \( \Delta \) is:
\[ \Delta = -8 \]